Question: Suppose we wanted to evaluate the double integral $S = \iint_D \ln(xy) \, dx \, dy$ by first applying a change of variables from $D$ to $R$ : $\begin{aligned} x &= X_1(u, v) = e^{uv} \\ \\ y &= X_2(u, v) = e^{u}e^{v} \end{aligned}$ What is $S$ under the change of variables? Assume $v > u$. If you know an expression within absolute value is non-negative, do not use absolute value at all. $S = \iint_R $ $du \, dv$
Answer: If we have a transformation $\bold{X} : R \to D$, then we can rewrite an integral under the change of variables: $ \iint_D f(x, y) \, dA = \iint_R f(\bold{X}(u, v)) | J(\bold{X}) | \, du \, dv$ First we need to find the absolute value of the Jacobian, $|J(\bold{X})|$. $\begin{aligned} |J(\bold{X})| &= \left| \det \begin{pmatrix} \dfrac{\partial X_1}{\partial u} & \dfrac{\partial X_1}{\partial v} \\ \\ \dfrac{\partial X_2}{\partial u} & \dfrac{\partial X_2}{\partial v} \end{pmatrix} \right| \\ \\ &= \left| \det \begin{pmatrix} ve^{uv} & ue^{uv} \\ \\ e^ue^v & e^ue^v \end{pmatrix} \right| \\ \\ &= \left| ve^{uv}e^ue^v - ue^{uv}e^ue^v\right| \\ \\ &= \left| e^{uv + u + v} (v - u) \right| \\ \\ &= e^{uv + u + v} \left| v - u \right| \end{aligned}$ We are given that $v > u$, so $e^{uv + u + v} \left| v - u \right|$ simplifies to $e^{uv + u + v} (v - u)$. Now we substitute $u$ and $v$ in $f(x, y)$. $\begin{aligned} f(x, y) &= f(\bold{X}(u, v)) \\ \\ &= \ln(X_1(u, v) X_2(u, v)) \\ \\ &= \ln( e^{uv} e^ue^v ) \\ \\ &= \ln(e^{uv + u + v}) \\ \\ &= uv + u + v \end{aligned}$ Putting everything together, we get the integral under the change of variables: $ \iint_R e^{uv + u + v} (v - u) (uv + u + v) \, du \, dv$